Application of Derivatives

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We all give exams in our school and college life. It is an inevitable part of our life.

After the exams, we always try to analyse our results and compare it with others.

What do we try to find out?

  • Who has got the highest (maximum) & who has got the lowest (minimum) marks?
  • What are the highest (maximum) & lowest (minimum) marks?
  • Is my own present performance better than the previous exam (increased) or lower than the previous one (decreased)?

Similarly, we will discuss some Applications of Derivatives to determine:

  1. The rate of change of quantities.
  2. Whether a function is increasing or decreasing.
  3. The maximum or minimum value of a function.

Note: If y = f(x) be a function then the derivative or the differential coefficient of f(x) with respect to x is denoted by  \frac{dy}{dx} or y’or f ‘ (x).

 

i.  The rate of change of quantities.

If y = f(x) be a function, the rate of change of a dependent variable (function ywith respect to independent variable (x) is given by

f ‘ (x)or \frac{dy}{dx} .

The rate of change of y with respect to (x) at x = a is denoted by f ‘ (a) or \left [ \frac{dy}{dx} \right ]_{x=a} .

 

Examples

Now let’s consider some examples on application of derivatives.

Example 1: Find the rate of change of area of a circle with respect to its radius r when its radius is 5 cm.

Let  be the area of a circle with radius cm.

Then A(r) =  \pir^2

\frac{dA}{dr} = A'(r) = 2\pi r

\left [ \frac{dA}{{dr}} \right ]_{r=5} = A'(5) = 2\pi(5) = 10\picm^2/cm

 

Example 2If C(x) = 5x2 + 10x + 9 represents the total cost (in dollars) of producing x items & the marginal cost is the rate of change of total cost with respect to the number of items produced. Find the marginal cost when 5 units are produced.

C(x) = 5x2 + 10x + 9

Marginal cost = C ‘(x) = \frac{dC}{dx} =10x+10+0=10x+10.

Marginal cost when 5 units are produced = \left [ \frac{dC}{dx} \right ]_{x=5}

\left [ \frac{dC}{dx} \right ]_{x=5}= 10(5)+10=50+10= $60.

 

ii.  Whether a function is increasing or decreasing.

If f(xis a function which is continuous on [ab& differentiable on (abthen –

(a) Increasing function: f(xis called increasing function on [ab], if f ‘ (x) > 0 for each x \in  (ab).

(b) Decreasing function: f(xis called decreasing function on [ab], if f ‘ (x) < 0 for each x \in  (ab).

(c) Constant function: f(xis called Constant function on [ab], if f ‘ (x) = 0 for each x \in  (ab).

 

Example 3Check whether the function f(x) = 12x + 17 is increasing on the set of real numbers i.e. .

Since f ‘ (x) = 12(1) + 0 = 12 > 0, for all \in  R.

Hence f(x) = 12x + 17 is increasing on the set of real numbers i.e. R.

 

Example 4Check whether the function f(x) = cos x is increasing or decreasing in \left ( 0, \frac{\pi}{2} \right ).

Since f ‘ (x) = sin x < 0 for all \in  \left ( 0, \frac{\pi}{2} \right ).

Hence f(x) = cos x is decreasing in \left ( 0, \frac{\pi}{2} \right ) .

 

iii.  The maximum or minimum value of a function.

If f(x) is a function which is defined on (ab), then f(x) is said to have:-

  • Maximum value at x = a if f(a) > f(x) for all x \in  (ab). Then f(a) is called the maximum value of f(x) in (ab) & x = a is called the point of maximum value or point of maxima.
  • Minimum value at x = a if f(a) < f(x) for all x \in  (ab). Then f(a) is called the minimum value of f(x) in (ab) and x = a is called the point of minimum value or point of minima.

First Derivative Test

  1. If f ‘ (xchanges sign from positive to negative as x increases through a, then x = a is a point of local maxima, and f(ais called the local maximum value.
  2. If f ‘ (xchanges sign from negative to positive as x increases through a, then x = a is a point of local minima, and f(ais called the local minimum value.
  3. If f ‘ (xdoes not change sign as x increases through a, then x = a is neither a point of local maxima nor a point of local minima. It is called a point of Inflection.

Example 5: Find all the points of local maxima & minima for the function, f(x) = 2x3 – 6x + 10.

Also find the local maximum & local minimum values.

f(x) = 2x3 – 6x + 10

f ‘ (x) = 6x2 – 6 = 6(x2 – 1) = 6(x + 1)(x – 1)

For maxima or minima, we first find the critical points.

Hence f ‘ (x) = 0 => 6(x + 1)(x – 1) = 0, which gives x = 1, -1.

Now, check x = 1, -1 using first derivative test for points of maxima or minima.

Observe that f(0.9) < 0 i.e. negative & f(1.1) > 0 i.e. positive.

Since f ‘ (xchanges sign from negative to positive as x increases through 1, therefore x = 1 is a point of local minima, and f(1) is called the local minimum value.

Local minimum value =f(1) = 2(1)3 – 6(1) + 10

Similarly, f ‘ (xchanges sign from positive to negative as x increases through -1, therefore x = -1 is a point of local maxima, and f(1) is called the local maximum value.

Local maximum value = f(-1) = 2(-1)3 – 6(-1) + 10 = -2 + 16 = 14

Check Point

  1. Find the rate of change of area of a circle with respect to its radius r when its radius is 15 cm.
  2. If R(x) = 5x3 + 2x + 7 represents the total revenue (in dollars) obtained from the sale of x units & the marginal revenue is the rate of change of total revenue with respect to the number of units sold. Find the marginal revenue when 10 units are sold.
  3. Check whether the function f(x) = 35x – 19 is increasing or decreasing on the set of real numbers R.
  4. Check whether the function f(x) = sin x is increasing or decreasing in \left ( 0, \frac{\pi}{2} \right ) .
  5. Find all the points of local maxima and minima for the function f(x) = x3 – 6x2 + 9x + 10.    Also, find the local maximum and local minimum values.
Answer Key
  1. 30\picm^2/cm
  2. Marginal Revenue = $1502.
  3. f(x) = 35x – 19 is increasing on the set of real numbers.
  4. f(x) = sin x is increasing in \left ( 0, \frac{\pi}{2} \right )
  5. Local maxima at x = 1, Local maximum value = 14

Local minima at x = 3, Local minimum value = 10

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